13.5 13.2: Calculus of Vector-Valued Functions

In this section, we extend differentiation and integration to vector-valued functions. This is straightforward because the techniques of single-variable calculus carry over with little change. What is new and important, however, is the geometric interpretation of the derivative as a tangent vector. We describe this later in the section.

The first step is to define limits of vector-valued functions.

DEFINITION Limit of a Vector-Valued Function

A vector-valued function r(t) approaches the limit u (a vector) as t approaches t0 if . In this case, we write

We can visualize the limit of a vector-valued function as a vector r(t) “moving” toward the limit vector u (Figure 1). According to the next theorem, vector limits may be computed componentwise.

The vector-valued function r(t) approaches u as t → t0.

THEOREM 1: Vector-Valued Limits Are Computed Componentwise

A vector-valued function r(t) = 〈x(t), y(t), z(t)〉 approaches a limit as t → t0 if and only if each component approaches a limit, and in this case,

The Limit Laws of scalar functions remain valid in the vector-valued case. They are verified by applying the Limit Laws to the components.

Proof

Let u = 〈a, b, c〉 and consider the square of the length

The term on the left approaches zero if and only if each term on the right approaches zero (because these terms are nonnegative). It follows that ∥r(t − u∥) approaches zero if and only if |x(t) − a|, |y(t) − b|, and |z(t) − c| tend to zero. Therefore, r(t) approaches a limit u as t → t0 if and only if x(t), y(t), and z(t) converge to the components a, b, and c.

EXAMPLE 1

Calculate , where r(t) = 〈t2, 1 − t, t-1〉.

Solution By Theorem 1,

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Continuity of vector-valued functions is defined in the same way as in the scalar case. A vector-valued function r(t) = 〈x(t), y(t), z(t)〉 is continuous at t0 if

By Theorem 1, r(t) is continuous at t0 if and only if the components x(t), y(t), z(t) are continuous at t0.

We define the derivative of r(t) as the limit of the difference quotient:

In Leibniz notation, the derivative is written dr/dt.

We say that r(t) is differentiable at t if the limit in Eq. (3) exists. Notice that the components of the difference quotient are difference quotients:

and by Theorem 1, r(t) is differentiable if and only if the components are differentiable. In this case, r′(t) is equal to the vector of derivatives 〈x′(t), y′(t), z′(t)〉.

By Theorems 1 and 2, vector-valued limits and derivatives are computed “componentwise,” so they are not more difficult to compute than ordinary limits and derivatives.

THEOREM 2: Vector-Valued Derivatives Are Computed Componentwise

A vector-valued function r(t) = 〈x(t), y(t), z(t)〉 is differentiable if and only if each component is differentiable. In this case,

Here are some vector-valued derivatives, computed componentwise:

Higher-order derivatives are defined by repeated differentiation:

EXAMPLE 2

Calculate r″(3), where r(t) = 〈ln t, t, t2〉.

Solution We perform the differentiation componentwise:

Therefore, .

The differentiation rules of single-variable calculus carry over to the vector setting.

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Differential Rules

Assume that r(t), r1(t), and r2(t) are differentiable. Then

  • Sum Rule:

  • Constant Multiple Rule: For any constant c, (c r(t))′ = c r′(t).

  • Product Rule: For any differentiable scalar-valued function f(t),

  • Chain Rule: For any differentiable scalar-valued function g(t),

Proof

Each rule is proved by applying the differentiation rules to the components. For example, to prove the Product Rule (we consider vector-valued functions in the plane, to keep the notation simple), we write

f(t)r(t) = f(t) 〈x(t), y(t)〉 = 〈f(t)x(t), f(t)y(t)〉

Now apply the Product Rule to each component:

The remaining proofs are left as exercises (Exercises 69–70).

EXAMPLE 3

Let r(t) = 〈t2, 5t, 1〉 and f(t) = e3t. Calculate:

Solution We have r′(t) = 〈2t, 5, 0〉 and f′(t) = 3e3t.

  • By the Product Rule,

  • By the Chain Rule,

There are three different Product Rules for vector-valued functions. In addition to the rule for the product of a scalar function f(t) and a vector-valued function r(t) stated above, there are Product Rules for the dot and cross products. These rules are very important in applications, as we will see.

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THEOREM 3: Product Rule for Dot and Cross Products

Assume that r1(t) and r2(t) are differentiable. Then

CAUTION

Order is important in the Product Rule for cross products. The first term in Eq. (5) must be written as

not . Similarly, the second term is . Why is order not a concern for dot products?

Proof

We verify Eq. (4) for vector-valued functions in the plane. If r1(t) = 〈x1(t), y1(t)〉 and r2(t) = 〈x2(t), y2(t)〉, then

The proof of Eq. (5) is left as an exercise (Exercise 71).

In the next example and throughout this chapter, all vector-valued functions are assumed differentiable, unless otherwise stated.

EXAMPLE 4

Prove the formula .

Solution By the Product Formula for cross products,

Here, r′(t) × r′(t) = 0 because the cross product of a vector with itself is zero.